InterviewSolution
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InterviewSolution
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A smooth pulley A of mass M_0 is lying on a smooth table. A light string passes round the pulley and has masses M_1 and M_2attached to its ends , the two portion of the string being perpendicular to the edge of the table so that the masses hang vertically . Calculate the acceleration of the pulley . |
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Answer» Solution :Let the length of portions of string on the table be `l_0 , l_1 and l_2` as shown in figure . Let mass `M_0` MOVE to the right by x on the table , `M_1` goes down by `y_1 and M_2` goes up by `y_2` . Then `l_0` becomes `(l_0 -x),l_1` becomes ` (l_1 + y_1) and l_2` becomes ` (l_2 - y_2)`. As the length of string remains unchanged ` 2l_0 + l_1 + l_2 = 2(l_0-x) + l_1 + y_1 + l_2 - y_2` or ` 2 x = y_1 -y_2` Diff. twice w.r. to t, we get ` 2(d^(2)x)/(dt^(2)) = (d^(2) y_1)/(dt^(2)) - (d^(2)y_2)/(dt^(2))` If ` a_0 , a_1 and a_2` are acceleration of  `M_0 , M_1 and M_2 ` respectively , then ` 2a_0 = a_1 - a_2` ......... (1) For motion of `M_0 , 2T= M_0a_0` ....... (2) `M_1g-T=M_1 a_1`....... (3) For motion of mass `M_2` , ` T-m_2g = M_2a_2` .......... (4) Substituting values of ` a_0 , a_1 and a_2` from (2) , (3) and (4) in equation (1) , we get ` 2((2T)/(M_0))=(g-(T)/(M_1)) -((T)/(M_2)-g)`, `(4T)/(M_0)= 2g-T((1)/(M_1) + (1)/(M_2)) i.e., ((4)/(M_0) + (1)/(M_1) + (1)/(M_2))T=2g ` This GIVES `T=(2M_0 M_1 M_2g)/(4M_1M_2+ M_0 (M_1 + M_2))`........ (5) `:.` Acceleration of puelly A from (2) `a_0 = (2T)/(M_0) = (4M_1M_2 g)/( 4M_1 M_2 + M_0(M_1 + M_2))` ............ (6) |
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